python multiprocessing BaseManager注册的类在Ctrl-C之后立即失去连接

Question

我遇到了一些问题，我怀疑这是我的python程序无法正确处理的限制，我的程序无法在我点击Ctrl-C后立即调用BaseManager的注册类的方法，甚至是其他作为从多处理继承的类实现的进程。流程受到影响。我有一些方法想从Ctrl-C之后不能正确执行的进程中调用。

例如，以下代码无法在Ctrl-C之后调用TestClass的mt实例。

from multiprocessing.managers import BaseManager, NamespaceProxy
import time
class TestClass(object):
    def __init__(self, a):
        self.a = a
    def b(self):
        print self.a
class MyManager(BaseManager): pass
class TestProxy(NamespaceProxy):
    # We need to expose the same __dunder__ methods as NamespaceProxy,
    # in addition to the b method.
    _exposed_ = ('__getattribute__', '__setattr__', '__delattr__', 'b')
    def b(self):
        callmethod = object.__getattribute__(self, '_callmethod')
        return callmethod('b')
MyManager.register('TestClass', TestClass, TestProxy)
if __name__ == '__main__':
    manager = MyManager()
    manager.start()
    t = TestClass(1)
    print t.a
    mt = manager.TestClass(2)
    print mt.a
    mt.a = 5
    mt.b()
    try:
        while 1:
            pass
    except (KeyboardInterrupt, SystemExit):
        time.sleep(0.1)
        mt.a = 7
        mt.b()
        print "bye"
        pass
Here is the console output
1
2
5
^CTraceback (most recent call last):
  File "testManager.py", line 38, in <module>
    mt.a = 7
  File "/usr/lib/python2.7/multiprocessing/managers.py", line 1028, in __setattr__
    return callmethod('__setattr__', (key, value))
  File "/usr/lib/python2.7/multiprocessing/managers.py", line 758, in _callmethod
    conn.send((self._id, methodname, args, kwds))
IOError: [Errno 32] Broken pipe

你有什么建议吗？我的代码中有什么变通方法或错误吗？

提前谢谢。

Answer 1

如果有人碰巧遇到这个问题，我会根据这个答案解决https://stackoverflow.com/a/21106459/1667319。这是的工作代码

from multiprocessing.managers import SyncManager, NamespaceProxy
import time
import signal
#handle SIGINT from SyncManager object
def mgr_sig_handler(signal, frame):
    print 'not closing the mgr'
#initilizer for SyncManager
def mgr_init():
    signal.signal(signal.SIGINT, mgr_sig_handler)
    #signal.signal(signal.SIGINT, signal.SIG_IGN) # <- OR do this to just ignore the signal
    print 'initialized mananger'
class TestClass(object):
    def __init__(self, a):
        self.a = a
    def b(self):
        print self.a
class MyManager(SyncManager): pass
class TestProxy(NamespaceProxy):
    # We need to expose the same __dunder__ methods as NamespaceProxy,
    # in addition to the b method.
    _exposed_ = ('__getattribute__', '__setattr__', '__delattr__', 'b')
    def b(self):
        callmethod = object.__getattribute__(self, '_callmethod')
        return callmethod('b')
MyManager.register('TestClass', TestClass, TestProxy)
if __name__ == '__main__':
    manager = MyManager()
    manager.start(mgr_init)
    t = TestClass(1)
    print t.a
    mt = manager.TestClass(2)
    print mt.a
    mt.a = 5
    mt.b()
    try:
        while 1:
            pass
    except (KeyboardInterrupt, SystemExit):
        time.sleep(0.1)
        mt.a = 7
        mt.b()
        print "bye"
        pass

干杯，

Answer 2

这是问题的重复，我在那里写道：

最简单的解决方案-使用启动管理器

manager.start(signal.signal, (signal.SIGINT, signal.SIG_IGN))

而不是manager.start（）。并检查信号模块是否在您的导入中（导入信号）。

这个捕获并忽略管理器进程中的SIGINT（Ctrl-C）。